Proving prime numbers

  • Thread starter l-1j-cho
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  • #1
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how do you prove, if p is prime, then a derived equation of p is prime, if true?
 

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  • #2
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Hi l-1j-cho! :smile:

What exactly do you mean with "a derived equation of p"?
 
  • #3
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hello!

uhm, I don't know anything in particular, but something like
if p is prime, the following equation is prime
or if p is prime, (2^p -1) is prime such things
 
  • #4
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hello!

uhm, I don't know anything in particular, but something like
if p is prime, the following equation is prime
or if p is prime, (2^p -1) is prime such things
You can't be saying "if p is prime, (2^p - 1) is prime" because that is not true for p = 11. But it is easy to show that (2^p -1) can be prime only if p is prime. As for equations being prime, do you mean proving that a polynominal is not factorable into the product of two polynominals of lower degree? Or do you mean to show that a certain polynomial in n yields primes for all positive values of n less than a certain integer?
 
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  • #5
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oh right my apology

I mean polynomials that is expressed in terms of p.
obviously, polynomials like p^2+5p+6 is not prime because if can be factored to (p+2)(p+3)
but my question is, how do you prove that a random polynomials always spits out a prime number whenever we plug in a prime number?

but before that, would such polynomial exist?
(not necesarrily polynomials but exponents or other stuff)
 
  • #6
841
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oh right my apology

I mean polynomials that is expressed in terms of p.
obviously, polynomials like p^2+5p+6 is not prime because if can be factored to (p+2)(p+3)
but my question is, how do you prove that a random polynomials always spits out a prime number whenever we plug in a prime number?

but before that, would such polynomial exist?
(not necesarrily polynomials but exponents or other stuff)
It has been proven that such polynominals don't exist since if P(n) is prime then P(n+a*prime) is also divisible by "prime",
. Don't know but very much doubt that there is some exotic function in P that is prime for all prime P.
 
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  • #7
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thank you!
 
  • #8
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P(n+a*prime) is also divisible by prime,
QUOTE]

sorry but, could you explain about this more?
 
  • #9
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It has been proven that such polynominals don't exist since if P(n) is prime then P(n+a*prime) is also divisible by "prime",
. Don't know but very much doubt that there is some exotic function in P that is prime for all prime P.

The fun thing that there IS such a function! The function is totally useless, but it exists: http://www.math.hmc.edu/funfacts/ffiles/10003.5.shtml
 
  • #10
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P(n+a*prime) is also divisible by prime,
QUOTE]

sorry but, could you explain about this more?

Sure 3p +2 = 23 for p = 7, so for p = 7 + a*23 e.g. for p = 7,30,53,... ,3p + 2 is divisible by 23 and hence not prime. The same goes for any polynominal in p.
 

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